Multiple choice question for engineering
1. Which of the following is not a type of arc welding?
Answer: d [Reason:] Among the following, plasma arc, electro-slag and submerged arc are few of the types of arc welding. Air-acetylene is one of the forms of gas welding processes.
2. Which of the following is not a type of resistance welding?
Answer: c [Reason:] Among the following, seam, projection and spot are few of the types of resistance welding. Electro-slag is one of the forms of gas welding processes.
3. Which of the following materials are not suited for cold welding?
Answer: b [Reason:] Cold welding process is not very suitable for alloys like steel, the reason being, cold welding cannot take place excellently in ferrous materials. Steel contains high amount of iron, hence it cannot be cold welded.
4. What is the cross-sectional shape of a thermoplastic welding rod?
Answer: a [Reason:] A welding rod or a thermoplastic welding rod is used mainly in plastic welding. The rod should not be very porous or it may catch voids. This rod has a cross sectional shape triangular or circular.
5. Which of the following is not a type of related processes?
Answer: b [Reason:] Among the following, soldering, pressing and hard-pressing are few of the types of related processes. Percussion is one of the forms of resistance welding processes.
6. Laser is classified under newer beam.
Answer: a [Reason:] There are various types of welding processes, one of them is newer welding process. Under this, there are two types, one is electron-beam and another is the use of laser.
7. Which of the following is not a type of solid state welding?
Answer: a [Reason:] Among the following, ultrasonic, friction and diffusion are a few of the types of related processes. Percussion is one of the forms of resistance welding processes.
8. What is the density of the plastic used in speed tip welding?
a) 916 kg/m3
b) 923 kg/m3
c) 946 kg/m3
d) 982 kg/m3
Answer: c [Reason:] The plastic which is used in the process of speed tip welding is polypropylene or polypropene. It is classified under thermoplastics materials. The density of polypropylene is 946 kg/m3 and it starts melting at 130°C.
9. Which of the following could be considered as a related process?
b) Oxy-Hydrogen welding
c) Oxy-Acetylene cutting
Answer: c [Reason:] Among the following, oxy-acetylene cutting is considered as a related process. Explosive is a type of solid state welding. Electron-beam is a type of newer welding and oxy-hydrogen welding is a form of gas welding.
10. A liquid phase has to be used in cold welding.
Answer: b [Reason:] It is hot welding that a liquid or a molten state is used for the joining of two pieces of metal. But in cold welding, there is no use of a liquid phase, that is, joining is done without using heat.
1. What is the capacity of automatic welding machine?
a) 100 to 200A
b) 300 to 400A
c) 500 to 700A
d) 800 to 3000A
Answer: d [Reason:] An automatic welding machine has a high amount of capacity. Its capacity ranges in between 800 to 300A. This type of a machine is not portable and is suitably used for heavy welding processes.
2. What is the capacity of a light manual welding machine?
a) 100 to 200A
b) 300 to 400A
c) 500 to 700A
d) 800 to 3000A
Answer: a [Reason:] A light manual welding machine has the one of the lowest capacity among all types of welding machines. Its capacity ranges in between 100 to 200A. This machine is portable and is best suited for light welding processes.
3. What is the current required in an Oxygen’s INDARC 400 (S) transformer?
a) 20 amp to 100 amp
b) 40 amp to 250 amp
c) 50 amp to 400 amp
d) 60 amp to 500 amp
Answer: c [Reason:] In an Oxygen’s INDARC 400 (S) transformer, there is a hand wheel provided which is used for varying the output, for altering the air gap in the core and results in regulating the current in between 50 amperes to 400 amperes.
4. What is the efficiency of an A.C. welding transformer?
Answer: c [Reason:] A.C welding is currently gaining considerable ground. The efficiency of an alternating current welding transformer varies from 0.8 to 0.85 and the efficiency of A.C. outfits is in the range of 0.3 to 0.6.
5. What is the amount of electric energy required per kg in dc welding operation?
a) 1-4 kWh
b) 3-7 kWh
c) 4-9 kWh
d) 6-10 kWh
Answer: d [Reason:] The amount of electrical energy consumed required per kg for a D.C. welding operation, is 6 kWh to 10 kWh. Direct current provides the advantage of distribution of the heat needed inside the arc.
6. Which material is not used for making non-consumable electrodes?
Answer: c [Reason:] Non-consumable electrodes, are those electrodes which do not get consumed as the welding operation proceeds. These electrodes are primarily made up of materials like carbon, graphite and tungsten.
7. Which of the following characteristic does not define a coated electrode?
a) Maintenance of arc
b) Protection from oxygen
c) Protection from nitrogen
d) Preventing other elements into molten metal
Answer: d [Reason:] A coated electrode is a further classification of a consumable electrode. The purpose of the electrode is to maintain the arc, to shield the molten metal from oxygen and nitrogen present in the air and to provide other alloying elements into the molten metal.
8. What is the amount of carbon contained by a soft steel wire?
a) 0.1% to 0.18%
b) 0.15% to 0.22%
c) 0.18% to 0.26%
d) 0.2% to 0.27%
Answer: a [Reason:] A soft steel wire contains about 0.1% to 0.18% of carbon. Soft steel wire also has other contents in it like phosphorous and sulphur in it. The amount of phosphorous is soft steel is around 0.0025%.
9. What is the percentage of carbon present in low alloy steel?
Answer: c [Reason:] The amount of carbon that is present in a soft steel wire is around 0.1% to 0.18% and the amount of carbon in terms of percentage composition in low alloy steels is 0.25%.
10. The presence of arc blower makes good quality welds.
Answer: b [Reason:] A flow of large amount of current causes the generation of arc blower. The space around the arc and the metal always experiences magnetic fields, causing the currents to get squared. This results in an unstable behavior by the arc blower, which is therefore not desirable for good welds.
1. Which of the following methods cannot be used for welding carbon steel?
a) Arc welding
b) Gas welding
c) Ultrasonic welding
d) Forge welding
Answer: c [Reason:] For the welding of carbon steel material, one may use the arc welding method. To gas weld carbon steel is also a feasible option and carbon steels can be treated using forge welding also, but it cannot be welded using ultrasonic welding.
2. High alloys steels are preheated up to what temperature?
Answer: b [Reason:] High alloy steels are those steels which contain high amount of carbon in them. These alloys need to be preheated before welding in order to avoid cracking. The temperature up to which these alloys are preheated is 400°C.
3. Up to what percent of carbon content in steel, it is not required to preheat it?
a) 0.25% – 0.3%
b) 0.3% – 0.35%
c) 0.35% – 0.4%
d) 0.4% – 0.45%
Answer: a [Reason:] Low carbon steels are the ones containing low amount of carbon in them, which does not require preheating. The amount of carbon content needed in low carbon steels, is 0.25% to 0.3%, if preheating is to be avoided.
4. Which of the following material is not used in alloy steels?
Answer: d [Reason:] In alloy steels, there is a big composition used up by iron. Along with iron, there vitally has to be carbon in alloy steels. Along with carbon, there is a presence of other elements such as, molybdenum, nickel and chromium in small amounts.
5. Which among the following methods is best suited for stainless steels?
a) Electric butt welding
b) Seam welding
c) Flux coated arc welding
d) Oxy-hydrogen welding
Answer: a [Reason:] There are a few methods of treating stainless steels for welding, such as, any of the metal arc welding methods or the oxy-acetylene welding method, but the one best suited for this alloy is electric butt welding.
6. Stainless steels are annealed at what temperature?
a) 650°C – 700°C
b) 700°C – 750°C
c) 750°C – 800°C
d) 800°C – 850°C
Answer: c [Reason:] Stainless steels are hardened by constant heating and cooling and the best method suited for this alloy is electric butt welding. As soon as the stainless steel alloy is electric butt welded, it needs to be annealed at a temperature in between 750°C to 800°C.
7. Which of the following is not a material used for making an electrode in metal arc welding, for treating stainless steels?
Answer: d [Reason:] Stainless steels are best welded using electric butt welding method and then followed by annealing at around 750°C to 800°C. Stainless steels can also be welded using the metal arc welding methods. For this the electrode which is needed, is made up of niobium, titanium and columbium. These materials avoid the occurrence of the defect called weld decay.
8. What is the temperature needed for preheating in cast irons?
Answer: b [Reason:] Cast irons, upon the removal of welding heat, get solidified very quickly. This solidification takes place due to the surrounding cold mass and the air around. Because of such quick cooling, the carbon gets retained and hard metal stays in the weld. To overcome this, cast irons have to be preheated to 600°C.
9. Low carbon steels need not be preheated before welding.
Answer: a [Reason:] Carbon steels which contain carbon in a very low amount are not compulsorily required to be preheated before undergoing welding, if the large runs are made. This is because, in low carbon steels, unlike high carbon steels, there is no cracking seen on welding.
10. Carbon steels can be fusion welded.
Answer: b [Reason:] Carbon steels are best treated using forge welding, resistance welding, arc welding or by gas welding. Carbon steels essentially cannot be fusion welded, because, of the occurrence of cracks which may cause due to carbon pick up.
1. The point form of Gauss law is given by, Div(V) = ρv
Answer: a [Reason:] The integral form of Gauss law is ∫∫∫ ρv dv = V. Thus differential or point form will be Div(V) = ρv.
2. If a function is said to be harmonic, then
a) Curl(Grad V) = 0
b) Div(Curl V) = 0
c) Div(Grad V) = 0
d) Grad(Curl V) = 0
Answer: c [Reason:] Though option a & b are also correct, for harmonic fields, the Laplacian of electric potential is zero. Now, Laplacian refers to Div(Grad V), which is zero for harmonic fields.
3. The Poisson equation cannot be determined from Laplace equation. State True/False.
Answer: b [Reason:] The Poisson equation is a general case for Laplace equation. If volume charge density exists for a field, then (Del)2V= -ρv/ε, which is called Poisson equation.
4. Given the potential V = 25 sin θ, in free space, determine whether V satisfies Laplace’s equation.
c) Data sufficient
d) Potential is not defined
Answer: a [Reason:] (Del)2V = 0
(Del)2V = (Del)2(25 sin θ), which is not equal to zero. Thus the field does not satisfy Laplace equation.
5. If a potential V is 2V at x = 1mm and is zero at x=0 and volume charge density is -106εo, constant throughout the free space region between x = 0 and x = 1mm. Calculate V at x = 0.5mm.
Answer: d [Reason:] Del2(V) = -ρv/εo= +106
On integrating twice with respect to x, V = 106. (x2/2) + C1x + C2.
Substitute the boundary conditions, x = 0, V = 0 and x = 1mm, V = 2V in V,
C1 = 1500 and C2 = 0. At x = 0.5mm, we get, V = 0.875V.
6. Find the Laplace equation value of the following potential field
V = x2 – y2 + z2
Answer: b [Reason:] (Del) V = 2x – 2y + 2z
(Del)2 V = 2 – 2 + 2= 2, which is non zero value. Thus it doesn’t satisfy Laplace equation.
7. Find the Laplace equation value of the following potential field
V = ρ cosφ + z
Answer: a [Reason:] (Del)2 (ρ cosφ + z)= (cos φ/r) – (cos φ/r) + 0
= 0, this satisfies Laplace equation. The value is 0.
8. Find the Laplace equation value of the following potential field
V = r cos θ + φ
Answer: d [Reason:] (Del)2 (r cos θ + φ) = (2 cosθ/r) – (2 cosθ/r) + 0
= 0, this satisfies Laplace equation. This value is 0.
9. The Laplacian operator cannot be used in which one the following?
a) Two dimensional heat equation
b) Two dimensional wave equation
c) Poisson equation
d) Maxwell equation
Answer: d [Reason:] The first three options are general cases of Laplacian equation. Maxwell equation uses only divergence and curl, which is first order differential equation, whereas Laplacian operator is second order differential equation. Thus Maxwell equation will not employ Laplacian operator.
10. When a potential satisfies Laplace equation, then it is said to be
Answer: d [Reason:] A field satisfying the Laplace equation is termed as harmonic field.
1. Find the conductivity of a material with conduction current density 100 units and electric field of 4 units.
Answer: a [Reason:] The conduction current density is given by, Jc = σE. To get conductivity, σ = J/E = 100/4 = 25 units.
2. Calculate the displacement current density when the electric flux density is 20sin 0.5t.
a) 10sin 0.5t
b) 10cos 0.5t
c) 20sin 2t
d) 20cos 2t
Answer: b [Reason:] The displacement current density is given by, Jd = dD/dt.
Jd = d(20sin 0.5t)/dt = 20cos 0.5t (0.5) = 10cos 0.5t.
3. Find the magnitude of the displacement current density in air at a frequency of 18GHz in frequency domain. Take electric field E as 4 units.
Answer: d [Reason:] Jd = dD/dt = εdE/dt in time domain. For frequency domain, convert using Fourier transform, Jd = εjωE. The magnitude of
Jd = εωE = ε(2πf)E. On substituting, we get 4 ampere.
4. Calculate the frequency at which the conduction and displacement currents become equal with unity conductivity in a material of permittivity 2.
a) 18 GHz
b) 9 GHz
c) 36 GHz
d) 24 GHz
Answer: b [Reason:] When Jd = Jc , we get εωE = σE. Thus εo(2∏f) = σ. On substituting conductivity as one and permittivity as 2, we get f = 9GHz.
5. The ratio of conduction to displacement current density is referred to as
a) Attenuation constant
b) Propagation constant
c) Loss tangent
d) Dielectric constant
Answer: c [Reason:] Jc /Jd is a standard ratio, which is referred to as loss tangent given by σ /ε ω. The loss tangent is used to determine if the material is a conductor or dielectric.
6. If the loss tangent is very less, then the material will be a
b) Lossless dielectric
c) Lossy dielectric
Answer: b [Reason:] If loss tangent is less, then σ /ε ω <<1. This implies the conductivity is very poor and the material should be a dielectric. Since it is specifically mentioned very less, assuming the conductivity to be zero, the dielectric will be lossless (ideal).
7. In good conductors, the electric and magnetic fields will be
a) 45 in phase
b) 45 out of phase
c) 90 in phase
d) 90 out of phase
Answer: b [Reason:] The electric and magnetic fields will be out of phase by 45 in good conductors. This is because their intrinsic impedance is given by η = √(ωμ/σ) X (1+j). In polar form we get 45 out of phase.
8. In free space, which of the following will be zero?
Answer: c [Reason:] In free space, ε = ε0 and μ = μ0. The relative permittivity and permeability will be unity. Since the free space will contain no charges in it, the conductivity will be zero.
9. If the intrinsic angle is 20, then find the loss tangent.
a) tan 20
b) tan 40
c) tan 60
d) tan 80
Answer: b [Reason:] The loss tangent is given by tan 2θn, where θn = 20. Thus the loss tangent will be tan 40.
10. The intrinsic impedance of free space is given by
a) 272 ohm
b) 412 ohm
c) 740 ohm
d) 377 ohm
Answer: d [Reason:] The intrinsic impedance is given by η = √(μo/εo) ohm. Here εo = 8.854 x 10-12 and μo = 4π x 10-7.
On substituting the values, we get η = 377 ohm.